It’s okay to learn here, just do the problem directly.
First, clarify the four functions of the switch.
- Forward different network frames
- Drop the same network frame
- Learning source address
- Broadcast unknown frames
After A->C, D->A, F->D, the
source address is stored in the forwarding table. The interface is as shown in the figure.
A->C has nothing, so it broadcasts. When S1 and S2 are both broadcast operations
D->A, S1 and S2 have already got the address of A, so S1 performs the broadcast operation.
F->D has nothing to do with S2 , S1 has the address of D, so broadcast, S2 did not operate in the previous step, so the broadcast is executed. The 
hub divides the bandwidth equally, and the switch does not
exchange isolation conflict domains, and does not isolate broadcast domains
Network layer
If 128 is divided into 2^7, the host number is 16-bit 16-7=9, and the
IP must be subtracted from the broadcast address and the host address
2^9-2=510 to
form a full array, as shown in the figure. Take the 5 digits in the third segment as the network number.
Start with 19 00100000, which is 32. So choose 
the frame delivered by the introduction of c . The relationship between the target and the source and the router depends on the previous PT experiment.
Fragmentation
MF (more fragment), DF (Dont fragment)
DF=0 can be fragmented, MF=1 means there are more fragments, MF=0 means there is no more fragment
It's too simple, that is, direct delivery and indirect delivery.
Pay attention to the MAC of the time frame, or the packet